Question: The grades on a physics midterm at Almond are normally distributed with $\mu = 70$ and $\sigma = 4.5$. Tiffany earned a $59$ on the exam. Find the z-score for Tiffany's exam grade. Round to two decimal places.
Solution: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Tiffany's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{59 - {70}}{{4.5}}} $ ${ z \approx -2.44}$ The z-score is $-2.44$. In other words, Tiffany's score was $2.44$ standard deviations below the mean.